This is similar to some other equations which have been solved here, but the "2x" is throwing me, I'm not sure what to do with it.The equation is 4^x-1 = 3^2x, solve for x. I have the answer, but have not been able to reproduce the steps to get to it.log4/(log4)-(2log3), then it is completed on a calculator.

Yes, I know how to take the log and expand it, but I am having trouble solving it for x. It seems that everything I do to solve for (x-1) messes up (2x) and vice versa. I'm probably overlooking something obvious.

danlers wrote:Yes, I know how to take the log and expand it, but I am having trouble solving it for x. It seems that everything I do to solve for (x-1) messes up (2x) and vice versa. I'm probably overlooking something obvious.

I've tried to solve this so many ways I don't even know what work to show. Let me ask a specific question. If I start by taking the log of both sides, should I do log (base 4), log (base 3), or log (base 10)?

Hmm, I see that. Apparently my problem is further on.Here's one of the ways I've tried it.4^(x-1) = 3^(2x)log4^(x-1) = log3^(2x)x(log4)-1(log4) = 2x(log3)x(log4)=2x(log3)+(log4)x/2x=(log3)+(log4)/log4and if I go a step further I would end up withx/x=2(log3)+(log4)/log4and x/x equals one. I always end up with some variation of this, where the x's cancel each other out. I can't get a single x isolated.Can you point me in the right direction?Thanks!